Q8.    Find the equation of the curve passing through the point \left(0,\frac{\pi}{4} \right ) whose differential equation is \sin x \cos y dx + \cos x \sin y dy = 0.

Answers (1)
G Gautam harsolia

Given equation is
\sin x \cos y dx + \cos x \sin y dy = 0.
we can rewrite it as
\frac{dy}{dx}= -\tan x\cot y\\ \\ \frac{dy}{\cot y}= -\tan xdx\\ \\ \tan y dy =- \tan x dx
Integrate both the sides
\log |\sec y|+C' = -\log|sec x|- C''\\ \\ \log|\sec y | +\log|\sec x| = C\\ \\ \sec y .\sec x = e^{C}
Now by using boundary conditiond, we will find the value of C
It is given that the curve passing through the point \left(0,\frac{\pi}{4} \right )
So,
\sec \frac{\pi}{4} .\sec 0 = e^{C}\\ \\ \sqrt2.1= e^C\\ \\ C = \log \sqrt2
Now,
\sec y.\sec x= e^{\log \sqrt 2}\\ \\ \frac{\sec x}{\cos y} = \sqrt 2\\ \\ \cos y = \frac{\sec x}{\sqrt 2}
Therefore, the equation of the curve passing through the point \left(0,\frac{\pi}{4} \right ) whose differential equation is \sin x \cos y dx + \cos x \sin y dy = 0. is   \cos y = \frac{\sec x}{\sqrt 2}

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